Problem 6

Determine the least possible value of

f(1998)

, where

f:\mathbb{N}\to\mathbb{N}

is a function such that for all

m,n\in\mathbb{N}

f\left(n^{2}f(m)\right)=m\left(f(n)\right)^{2}.

Step 2 of 6: Show that

k

divides every value

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In plain words

Compare prime valuations in the iterated identity.

k^r f(t^{r+1})=f(t)^{r+1}

Detailed analysis

An induction using $f(kt)=kf(t)$ and $f(kt^2)=f(t)^2$ gives $k^r f(t^{r+1})=f(t)^{r+1}$ for every $r\ge1$ . If a prime has valuation $a$ in $k$ and valuation $b<a$ in $f(t)$ , choosing $r$ with $ra>(r+1)b$ contradicts this identity. Therefore $k\mid f(t)$ for every $t$ .